MT【300】余弦的三倍角公式

2017清华大学THUSSAT附加学科测试数学(二测)
$\cos^5\dfrac{\pi}{9}+\cos^5\dfrac{5\pi}{9}+\cos^5\dfrac{7\pi}{9}$ 的值为_____
A.$\frac{15}{32}$

B.$\frac{15}{16}$

C.$\frac{8}{15}$

D.$\frac{16}{15}$

解答:注意到$\cos3\theta=4\cos^3\theta-3\cos\theta,\cos3\theta=\dfrac{1}{2}$的三个根为$\dfrac{\pi}{9},\dfrac{5\pi}{9},\dfrac{7\pi}{9}$故$\cos\dfrac{\pi}{9},\cos\dfrac{5\pi}{9},\cos\dfrac{7\pi}{9}$ 为$4\cos^3\theta-3\cos\theta-\dfrac{1}{2}=0$的三个根,即$\cos^3\theta=\dfrac{3}{4}\cos\theta+\dfrac{1}{8}$;
故$\cos^5\theta=\cos^2\theta\left(\dfrac{3}{4}\cos\theta+\dfrac{1}{8}\right)=\dfrac{3}{4}\cos^3\theta+\dfrac{1}{8}\cos^2\theta=\dfrac{9}{16}\cos\theta+\dfrac{3}{32}+\dfrac{1}{8}\cos^2\theta$
故$\cos^5\dfrac{\pi}{9}+\cos^5\dfrac{5\pi}{9}+\cos^5\dfrac{7\pi}{9}$

$=\dfrac{1}{8}(\cos\dfrac{\pi}{9}+\cos\dfrac{5\pi}{9}+\cos\dfrac{7\pi}{9})^2-\dfrac{1}{4}(\cos\dfrac{\pi}{9}\cdot\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{9}\cdot\cos\dfrac{7\pi}{9}+\cos\dfrac{7\pi}{9}\cdot\cos\dfrac{\pi}{9})+\dfrac{9}{32}$
$=\dfrac{1}{8}\cdot0^2-\dfrac{1}{4}\cdot(-\dfrac{3}{4})+\dfrac{9}{32}=\dfrac{15}{32}$

注:也可以用正余弦的快速降幂公式去做

注:一般的$\cos^n\dfrac{\pi}{9}+\cos^n\dfrac{3\pi}{9}\cos^n\dfrac{5\pi}{9}+\cos^n\dfrac{7\pi}{9}=\dfrac{1}{2}$

MT【300】余弦的三倍角公式

原文：https://www.cnblogs.com/mathstudy/p/10355544.html